Straight Line Motion (continued) Chapter 2 Straight Line Motion (continued)

Review Displacement: change in position Average velocity: rate of change in position over time 8/29/2019

Average Velocity Black line is the actual path Green line is the “average” path Not completely descriptive t x(t) Dx Dt t1 t2 x1 x2 8/29/2019

Ball Thrown in the Air “On average” it’s velocity is zero (Dx=0) We can do better by drawing two lines As we draw more lines, the approximation gets better 8/29/2019

Smaller and Smaller Dt The smaller Dt is, the better our “straight line” approximation looks Dt = 5 s Dt = 2.5 s Dt = 0.5 s 8/29/2019

Instantaneous Velocity What if we want to look at really small Dt? Look familiar? 8/29/2019

Position vs. Time Graph Velocity is equal to the slope of the tangent line to x(t) Positive (negative) velocity means the cart is moving in the positive (negative) direction v = 0 at a direction change 8/29/2019

Bungee Jumper Consider the velocity of a bungee jumper When is the velocity negative? When is the velocity positive When is the velocity zero? What does this tell us? 8/29/2019

x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2. Example You observe that a flea running on your ruler has a position given by x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2. What is the instantaneous velocity of the flea at t=2 s? At what time does it change direction? We know that velocity is the time derivative of position. The derivative is linear, so 8/29/2019

x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2 Solution Taking the derivative of x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2 Adding the terms together: 8/29/2019

Solution Finally, we plug in t = 2 s and solve: 8/29/2019

Solution At what time does the flea change direction? When v(t) = 0 8/29/2019

Displacement as an Integral of Velocity Consider a particle with a constant velocity Rearrange to give x(t2) Knowing v gives x(t) (almost – still need x(t1)) 8/29/2019

Displacement as an Integral of Velocity For constant velocity: Dx=v Dt The area under the curve is the displacement! t v(t) v Dt 8/29/2019

Displacement as an Integral of Velocity v(t) t Imagine breaking the a complex curve into lots of little rectangles This is just an integral! Displacement is the area under the curve 8/29/2019

Displacement as an Integral of Velocity We can arrive at the same result using calculus… 8/29/2019

Acceleration Time rate of change of velocity Units: m/s2 Average acceleration Instantaneous acceleration Units: m/s2 8/29/2019

Velocity vs. Time Graph (cont) Acceleration is equal to the tangent line to the slope of v(t) Note: An object can have a negative acceleration and still have a positive velocity! t v(t) Zero accel. Positive accel. Negative accel. 8/29/2019

Velocity as an Integral of Acceleration Same procedure as before… 8/29/2019

Constant Velocity Graph v(t) t Velocity vs Time x(t) t Position vs Time v(t) = constant x(t)=v t + x(0) 8/29/2019

Kinematics variables In one dimension Position x(t) meters Velocity v(t) meters/second Acceleration a(t) meters/second2 All depend on time All are vectors (have direction and magnitude) Position Velocity Acceleration d dt 6/17/04

Physical Interpretation of Zero x = 0 Object is at the origin v = 0 Object stopped or changing directions a = 0 Velocity is constant 8/29/2019

Special Case: Constant Acceleration In many cases, acceleration will be constant. We can use this to make our lives easier. Be careful: The following equations ONLY work if acceleration is constant. 8/29/2019

What Does It Look Like? Acceleration vs Time Velocity vs Time x t Position vs Time 8/29/2019

Constant Acceleration Given initial position and velocity, we can calculate position: Define: x(0) = x0 v(0) = v0 8/29/2019

Constant Acceleration We can use calculus to get this equation, too! 8/29/2019

Constant Acceleration So with constant acceleration, the fundamental equations are Should be able to solve any constant acceleration problem from these two, but… 8/29/2019

Constant Acceleration We can derive variants of these equations which are also useful Equation Missing Quantity vf = v0 + a t x – x0 x – x0 = v0 t + ½ a t2 vf vf 2 = v02 + 2 a (x – x0) t x – x0 = ½ (v0 + vf) t a x – x0 = vf t – ½ a t2 v0 8/29/2019

Free-Fall Acceleration Constant acceleration caused by gravity Only true if there is no air resistance We’ll ignore the effects of air resistance most of the time Near Earth’s surface, the acceleration due to gravity is approximately constant, g = 9.8 m/s2 DOWN This is a pretty good approximation 8/29/2019

If You Believe… A popular demonstration of free fall was conducted by Commander David Scott on the moon… 8/29/2019

Problem Solving Tips Read the problem carefully Set up the problem What facts are you given? What is asked for? Set up the problem Sketch a diagram Decide where x = 0, when t = 0 Decide which way is positive x direction Work from fundamental equations Don’t try to memorize every formula. Don’t just search the book for any formula containing the right symbols! 8/29/2019

Problem Solving Tips Reason symbolically as long as possible Calculator-punching errors are harder to find than algebra errors Use standard units Check result for correct units Practice builds skill and confidence 8/29/2019

Example: 150 ft It is rumored that Galileo dropped steel balls off the Tower of Pisa to determine that g was the same for all objects. How long does it take for a ball to reach the ground and how fast is it moving when it gets there? (neglect air resistance) 8/29/2019

The Solution 45.7 m Convert to standard units x=0 +x Convert to standard units 150 ft=45.7 m 2. Choose a coordinate system Origin: x=0 on ground Direction: up positive Drop begins at t=0 3. Write down the equation for position 8/29/2019

The Solution 45.7 m 4. Note that v0=0 5. Find t when x(t)=0 +x x=0 8/29/2019

The Solution x x=0 45.7 m 6. Solve for t with x0=45.7 m 7. Solve for v(t = 3.05 s) 8/29/2019

Question: 45.7 m x=0 x Let’s say Galileo dropped two steel balls. After dropping the first ball he waited one second and dropped the second ball How long after the first hits the ground does the second hit? Easy! 1 s 8/29/2019

The student is caught when xp = xs Example A stopped police car can accelerate at 30 m/s2. A PSU student passes the stopped police car going at the constant speed of vs = 50 m/s (110 mph). If the policeman takes 2 seconds to start his engine, how long does it take him to catch up to the student? Let xp be the position of the policeman and xs be the position of the student. The student is caught when xp = xs 8/29/2019

Solution: There is more than one way to “skin a cat” I think the easiest solution is to start the clock when the policeman begins accelerating, so we have the following: x0s=? v0p=0 m/s a0p=30 m/s2 v0s=50 m/s a0s=0 m/s2 +x Need to find x0s first 8/29/2019

Solution The car travels for 2 s at v = 50 m/s before we start the clock, so Now we can write down everyone’s position as a function of time This simplifies to: 8/29/2019

Solution Setting xp equal to xs: We have a quadratic equation. Recall: Solving equation 1 for t: 8/29/2019

Solution Simplifying and plugging in values Remember, this is 4.7 s after the engine was started, so it is 6.7 s after the student passed the policeman Why are there two answers, and why is one of them negative? 8/29/2019

Solution There are two solutions because the lines meet in two places The negative solution is non-physical x t t = 0 s 8/29/2019

One more Example How much acceleration must be provided by the brakes to stop a car traveling 30 m/s in 100 m? X=0 x Let x=0 m when braking begins. Then xf=100 m, vi=30 m/s, vf= 0 m/s 8/29/2019

Bouncing a Tennis ball Sketch x(t), v(t), a(t) for a basketball during one ‘dribble’. a -g bounce down up v0 v t t 8/29/2019

Bouncing a Tennis ball Sketch x(t), v(t), a(t) for a basketball during one ‘dribble’. bounce down up v0 v t x x0 t 8/29/2019